Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

public class Main  {

	

	static double[][] dp;
	static double[]arr;
	static int n;
	public static void main(String[] args)
	{
		Scanner sc=new Scanner(System.in);
	     n=sc.nextInt();
	    arr=new double[n];
	    for(int i=0;i<n;i++)
	    {
	    	arr[i]=sc.nextDouble();
	    	
	    }
	    dp=new double[n][n+1];
	    dp[0][0]=1-arr[0];
	    dp[0][1]=arr[0];
	    for(int i=1;i<n;i++)
	    {
	    	for(int j=0;j<=n;j++)
	    	{
	    		dp[i][j]=(1-arr[i])*dp[i-1][j];
	    		if(j>0)
	    		{
	    			dp[i][j]+=arr[i]*dp[i-1][j-1];
	    		}
	    		
	    	}
	    }
	    double ans=0;
	    for(int i=n/2+1;i<=n;i++)
	    {
	    	ans+=dp[n-1][i];
	    	
	    }
	    System.out.println(ans);
	}
	
	
	
}